Question

ABC is an equilateral triangle.Medians AD and CE are meeting ar point P . If AP=16CM. Find the perimeter of ∆ABC

Answer 1

Answer:

Step-by-step explanation:

GIVEN: Triangle ABC, 3 medians intersecting at G.

Area(triangle ABC) = 27 cm²

G is the centroid of the triangle, which intersects each medion in the ratio 2:1.

Here, AG : GM = 2:1

So, if AG = 2a, GM = a

TO FIND: area(triangle BCG)

CONSTRUCTION: CX perpendicular to AM

CALCULATION:

area(triangle CAG) = 1/2 * AG * CX =

1/2 * 2a *CX……..(1)

area(triangle CGM) = 1/2 * GM * CX =

1/2 * a * CX……….(2)

So area (triangle CGM) = 1/2 of area(triangle CAG) ……..(3)

Similarly, ar(tri BGM) = 1/2 of ar(tri BAG)….(4)

By adding (3) & (4)

ar(tri BCG) = 1/2 {ar(triCAG) + ar(triBAG)}

So, if ar(tri BCG) = A ………….(5)

Then ar(triCAG) + ar(triBAG) = 2A……….(6)

Adding (5) & (6)

We get ar(tri ABC ) = 3A = 27

So, A= 27/3 = 9

=> ar( tri BCG) = 9 cm²

Answer 2

Step-by-step explanation:

hello

I am solving it but at first you should know that median of equilateral triangle is perpendicular to the side on which it is.

also medians of a triangle divides each other in the ratio 2:1

So AP:DP =2:1 therefore DP=8 cm

therefore AP+DP =AD=24 cm

let the side of the equilateral triangle is 2a cm

therefore BD=a cm

In right angle triangle ADB (2a)^2=a^2+AD^2

where AD=24 cm

solving it we will get a=8 root 3 cm

therefore 2a =each side=16 root 3 cm

so perimeter will be 48 root 3cm (ans)

I will recommend to go throgh steps and check for calculation mistakes

HOPE IT HELPS YOU FOLLOW ME PLEASE!!!!!